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This problem involves using Gauss's Law to determine the electric field due to an infinite sheet of charge. Here’s the breakdown:

### Given Information:
- The infinite sheet has a **uniform surface charge density** \( \sigma \).
- A **cylindrical Gaussian surface** is used to apply Gauss’s Law, with its axis perpendicular to the sheet.
- Point \( P \) is located at the center of the top surface of the cylinder, which has an area \( A \) and is a distance \( D \) above the sheet.

### Gauss’s Law:
\[
\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0}
\]
Where:
- \( \mathbf{E} \) is the electric field.
- \( d\mathbf{A} \) is the area element.
- \( Q_{\text{enc}} \) is the charge enclosed by the Gaussian surface.
- \( \epsilon_0 \) is the permittivity of free space.

### Applying Gauss’s Law:
Since the Gaussian surface is symmetric about the sheet, the electric field will be uniform and perpendicular to the sheet. The total electric flux through the cylindrical Gaussian surface is the sum of the flux through the top and bottom surfaces of the cylinder.

1. **Electric flux**: The electric field penetrates both the top and bottom surfaces of the Gaussian surface. Thus, the total flux is:
\[
\Phi_E = E \cdot A + E \cdot A = 2EA
\]

2. **Enclosed charge**: The charge enclosed by the Gaussian surface is the surface charge density \( \sigma \) times the area \( A \) of the sheet:
\[
Q_{\text{enc}} = \sigma \cdot A
\]

3. **Gauss’s Law equation**:
\[
2EA = \frac{\sigma A}{\epsilon_0}
\]
Canceling out \( A \) from both sides:
\[
2E = \frac{\sigma}{\epsilon_0}
\]
So, the electric field magnitude \( E \) is:
\[
E = \frac{\sigma}{2\epsilon_0}
\]

### Answer:
From the options given, the correct equation is **(B)**. The equation matches the form:
\[
E \times 2A = \frac{\sigma A}{\epsilon_0}
\]

Would you like more details or have any further questions?

### Related Questions:
1. How is Gauss's Law applied to symmetric charge distributions like spheres or cylinders?
2. Why does the electric field from an infinite sheet of charge remain constant regardless of distance from the sheet?
3. What is the significance of using a Gaussian surface, and how is it chosen?
4. Can the electric field ever depend on the distance \( D \) from the sheet for a different surface charge configuration?
5. How would the answer change if the sheet were not infinite but finite?

### Tip:
When using Gauss's Law, always choose a Gaussian surface that matches the symmetry of the charge distribution—this simplifies the calculation of electric flux.

An infinite sheet has a uniform surface charge density σ. The figure shows a cylindrical Gaussian surface that is half above and half below the sheet, with its axis perpendicular to the sheet. Point P is located at the center of the top surface of the cylinder, which has area A and is a distance D above the sheet. Which of the following equations could be solved to determine the magnitude E of the electric field at Point P?

This problem involves applying Gauss's Law to determine the electric field at a point near an infinite sheet of charge with surface charge density σ. The problem uses a cylindrical Gaussian surface to find the electric field at Point P, located at a distance D from the sheet.

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